Abstract Algebra

Paul Melvin Bryn Mawr College

Fall 2011

lecture notes loosely based on Dummit and Foote’s text

Abstract Algebra (3rd ed)

Prerequisite:

Linear Algebra (203)

1

Introduction

Pure Mathematics

Algebra

Geom etry & Topology

Analysis

Foundations (set theory/logic)

What is Algebra?

• Number systems N = {1, 2, 3, . . . } “natural numbers”

Z = {. . . , −1, 0, 1, 2, . . . } “integers”

Q = {fractions} “rational numbers”

R = {decimals} = pts on the line “real numbers” √

C = {a + bi | a, b ∈ R, i = −1} = pts in the plane “complex nos”

k polar form

iθre , where a = r cos θ, b = r sin θ

√ Note N ⊂ Z ⊂ Q ⊂ R ⊂ C (all proper inclusions, e.g. 2 6∈ Q; exercise) There are many other important number systems inside C.

a

b a + bi

r θ

2

• Structure “binary operations” + and ·

associative, commutative, and distributive properties “identity elements” 0 and 1 for + and · resp.

solve equations, e.g. 1iax2 + bx + c = 0 has two (complex) solutions √

−b ± b2 − 4ac x =

2a

2 2 22i x + y = z has infinitely many solutions, even in N (the “Pythagorian triples”: (3,4,5), (5,12,13), . . . ).

n n n3i x + y = z has no solutions x, y, z ∈ N for any fixed n ≥ 3 (Fermat’s Last Theorem, proved in 1995 by Andrew Wiles; we’ll give a proof for n = 3 at end of semester).

• Abstract systems groups, rings, fields, vector spaces, modules, . . .

A group is a set G with an associative binary operation ∗ which has an identity element e (x ∗ e = x = e ∗ x for all x ∈ G) and inverses for each of its elements (∀ x ∈ G, ∃ y ∈ G such that x ∗ y = y ∗ x = e).

Examples (N, +) is not a group: no identity. (Z, +) is. (Z, ·) is not: no inverses. (Q, ·) isn’t, but (Q − {0}, ·) is.

Focus of first semester: groups and rings

Some history Theory of equations ( modern algebra)

• Quadratic equation (antiquity)

ax 2 + bx + c = 0

Divide by a and complete the square, i.e. substitute y = x + b/2a to get

y 2 + p = 0

where p = c/a − (b/2a)2 . The roots are y = ±√−p which gives the usual

formula by substitution.

3

2Remark Δ = b2 − 4ac = −4a p is called the discriminant of the polyno-mial ax2 + bx + c. If a, b, c ∈ R, then the equation has 2, 1 or 0 real roots according to the sign of Δ:

Δ > 0 Δ = 0 Δ < 0 0 0 iθ0Aside (complex mult and roots) If z = reiθ and z = r e , then

0 0 i(θ+θ0)zz = rr e ,

i.e. lengths multiply and angles add (prove this using trigonometry).∗

z0

z

r0

r θ0

θ

zz0

rr0

It follows that each nonzero complex number has two square roots, three cube roots, etc. In particular, the nth roots of z = reiθ are

u, uω, . . . , uωn−1

1/n iθ/n and ω 2πi/n where u = r e = e . These points are equally distributed on a circle of radius r1/n about the origin (since multiplication by ω rotates C about 0 by 2π/n radians).

• Cubic equation (16th century: del Ferro, Tartaglia Cardan, Viete)

ax 3 + bx2 + cx + d = 0

Divide by a and complete the cube, i.e. substitute y = x + b/3a to eliminate the quadratic term. This gives

y 3 + py + q = 0 ∗ Thus for any z, z 0 in the unit circle S1 , we have zz 0 , z −1 ∈ S1 so (S1 , ·) is a group!

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which has three roots y1, y2, y3. They can be found using Viete’s magical substitution

y = v(z) = z − p

.3z

3Indeed, this leads to a quadratic equation in w = z ,

w 2 + qw − (p/3)3 = 0 √

whose roots are w = −q/2 ± Δ, where Δ = (q/2)2 + (p/3)3 . Thus � √ �1/3 z = −q/2 ± Δ

(choose any one of the six possible cube roots) which yields Cardan’s formula

y1 = v(z) y2 = v(ωz) y3 = v(ω2 z) √

2πi/3where ω = e = −(1/2) + ( 3/2)i. The roots of the original equation are now obtained by subtracting b/3a.

Example Find the roots of x3 + 6x2 + 9x + 2.

Solution Substituting y = x + 2 gives y3 − 3y. Viete’s substitution is −1 3y = z + z , giving the quadratic w2 + 1 = 0 in w = z , with roots w = ±i.

Thus z is any cube root of ±i, say z = i. Cardan’s formula then gives √ √ y1 = i + i−1 = 0, y2 = ωi + (ωi)−1 = √− 3, y3 = ω√

2i + (ω2i)−1 = 3 (draw a picture), and so x1 = −2, x2 = − 3 − 2, x3 = 3 − 2 are the roots.

Exercise (not to hand in) Verify that y1, y2, y3 are indeed roots of the eqn. Hint: r is a root ⇐⇒ (y − r) is a factor of y3 + py + q, so it suffices to show y3 + py + q = (y − y1)(y − y2)(y − y3). But expanding out the last expression gives y3 − (y1 + y2 + y3)y2 + (y1y2 + y1y3 + y2y3)y − y1y2y3. So you must show 1i y1 + y2 + y3 = 0, 2i y1y2 + y1y3 + y2y3 = p and 3i y1y2y3 = q. Use the fact that ω3 = 1 and 1 + ω + ω2 = 0.

Question What is the significance of the sign of Δ?

• Quartic equation 16th century: Ferrari

• Quintic equation (and higher degree) 19th century: Abel proved that there does not exist a formula for the roots! Galois developed general theory (second semester) birth of modern algebra.

5

Some arithmetic (study of the natural numbers)

Definition Say d divides (or is a divisor of) a , written d|a, if ∃c with a = cd. Denote the greatest common divisor of a and b by gcd(a, b).

GCD Lemma ∀ a, b ∈ N, ∃ x, y ∈ Z such that gcd(a, b) = ax + by.

(e.g. gcd(10, 14) = 2 = 10 · 3 + 14 · (−2))

Proof Consider S = {ax + by | x, y ∈ Z}. (Remark S is closed under subtraction, and multiplication by any integer; we say S is an ideal in Z.) Set

d = min(S ∩ N)

the smallest positive elt of S. Claim d = gcd(a, b). Must show

1i d|a, d|b 2i e|a, e|b =⇒ e ≤ d

1i Clearly a = qd + r for some q, r ∈ Z w/ 0 ≤ r < d, so r = a − qd ∈ S (by the remark). By the minimality of d, r = 0, and so d|a. Similarly d|b. 2ie|a, e|b =⇒ e|(ax + by) for all x, y, and in particular e|d. Thus e ≤ d.

This proof used structural prop of Z (+, ·, <: Z is an “ordered ring”), the “Well Ordering Principle” (WOP) and the “Division Algorithm” (DA):

WOP Every non-empty subset of the natural numbers has a least element (or in symbols, S ⊂ N, S 6= ∅ =⇒ ∃ m ∈ S such that m ≤ s for all s ∈ S)

DA ∀a, d ∈ N, ∃q, r ∈ Z with 0 ≤ r < d such that a = qd + r

In fact can prove DA from WOP: Let S = {a − xd | x ∈ Z, a ≥ xd} ⊂ N ∪ {0}. By WOP, ∃r = min S = a − qd, i.e. a = qd + r, for some q. Then r < d. For if r ≥ d, then 0 ≤ r − d = a − (q + 1)d < r, contradicting the minimality of r.

Where does GCD lead? To Euclid’s Lemma and the

Fundamental Theorem of Arithmetic (FTA) Every natural no. n > 1 can be written as a product of primes. This product is unique up to the order of the factors.

(A natural number is prime if it has exactly two divisors)

6

Euclid’s Lemma If p is prime and p|ab, then p|a or p|b.

Proof Suppose p - a. Then gcd(a, p) = 1 (since p is prime) and so by the GCD Lemma, 1 = ax + py for some x, y. Thus b = 1 · b = (ax + py)b = (ab)x + pyb, which is clearly divisible by p.

Summarizing the logic so far: WOP =⇒ Euclid’s Lemma (via DA and GCD). For FTA, also need “induction”, which we use informally in the following proof (see below for more formal treatment).

Proof of FTA (existence) If n is prime, there’s nothing to prove. If not, then n = ab for some a, b < n. But then by induction, each of a, b has a prime decomposition. Put these together to get one for n. (uniqueness) Suppose p1 · · · pr = q1 · · · qs (all pi’s and qj ’s are prime).

Clearly p1|p1 · · · pr, so p1|q1q2 · · · qs. By Euclid’s Lemma and induction, p1

divides at least one of the qj ’s; can assume p1|q1 by reordering. But this implies p1 = q1 since q1 is prime. Thus p2 · · · pr = q2 · · · qs. The result follows by induction.

Induction

Principle of Induction If S ⊂ N satisfies

1i1 ∈ S and 2in ∈ S =⇒ n + 1 ∈ S

then S = N.

Theorem WOP ⇐⇒ Induction

=⇒ is HW #2. This shows that in fact WOP =⇒ FTA

Example Prove by induction on n that 1 + · · · + n = n(n + 1)/2.

Proof Let S = {k ∈ N | 1 + · · · + k = k(k + 1)/2}. Then 1 ∈ S, since 1 = 1(2)/2. Assume n ∈ S, and so

1 + · · · + n = n(n + 1)/2.

We must show n + 1 ∈ S. Adding n + 1 to both sides gives

1 + · · · + n + (n + 1) = n(n + 1)/2 + (n + 1) = (n(n + 1) + 2(n + 1))/2 = (n + 1)(n + 2)/2 = (n + 1)((n + 1) + 1)/2.

Hence n + 1 ∈ S, and so by induction, S = N.

7

I Foundations

§0. Sets

Assume familiarity with basics of set theory: set S = {· · · | · · · }

elements x ∈ S

subsets A ⊂ S

proper subset A $ S

union S ∪ T = {x | x ∈ S or x ∈ T }

intersection S ∩ T = {x | x ∈ S and x ∈ T }

difference S − T = {x | x ∈ S and x ∈/ T }

cartesian product S × T = {(s, t) | s ∈ S, t ∈ T }

cardinality |S| = # elements in S (if S is finite)

Notation: ∀, ∃, !, =⇒, ⇐⇒ (if and only if, or iff), =⇒⇐= (contradiction).

§1. Functions

fDefinition A function f : S → T (also written S → T ) consists of a

pair of sets S, T (the domain and codomain of the function) and a “rule” s 7→ f(s) assigning to each element s ∈ S an element f(s) ∈ T . (To be precise, a “rule” consists of a subset R ⊂ S × T satisfying ∀s ∈ S, ∃! t ∈ T such that (s, t) ∈ R, so a function is really a triple (S, T, R ⊂ S × T )...)

Remark The domain and codomain are essential parts of the function. For example the functions R → R and R → R+ (the real nos ≥ 0), both

2given by the rule x 7→ x , are distinct.

Examples 1i identity functions idS ) : S → S, s 7→ s.

2iinclusion of a subset A ⊂ S: A ,→ S, a 7→ a.

3irestriction of f : S → T to a subset A ⊂ S: f |A : A → T , a 7→ f(a).

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4iprojections S ← S × T → T , s ← (s, t) 7→ t.

5iconstant functions S → T , s 7→ t0, where t0 is a fixed elt of T .

6i composition of functions Given functions g : R → S and f : S → T , define f ◦ g : R → T by (f ◦ g)(r) = f(g(r)). Thus f ◦ g (also written fg) is defined by the “commutative diagram”

f ◦gR −→ T

g& %f

S

Definition Given f : S → T and subsets S0 ⊂ S and T0 ⊂ T , define the image of S0 under f to be

f(S0) := {f(s) | s ∈ S0} ⊂ T

and the preimage of T0 under f to be

f−1(T0) := {s ∈ S | f(s) ∈ T0} ⊂ S.

(examples and pictures) Call f(S) the image of f , also denoted im(f).

Proposition 1.1 aif−1(P ∪ Q) = f−1(P ) ∪ f−1(Q)

bif−1(P ∩ Q) = f−1(P ) ∩ f−1(Q)

Proof ai (Note: often prove = of sets in two steps, ⊂ and ⊃, though sometimes done simultaneously) x ∈ f−1(P ∪ Q) ⇐⇒ f(x) ∈ P ∪ Q ⇐⇒ f(x) ∈ P or f(x) ∈ Q ⇐⇒ x ∈ f−1(P ) or x ∈ f−1(Q) ⇐⇒ x ∈ f−1(P ) ∪ f−1(Q) bi iSame as a with ∪ and “or” replaced with ∩ and “and”.

Definition A function f : S → T is one-to-one (or monic or injective) if it maps at most one element of S to any given element of T . In other words, f(x) = f(y) =⇒ x = y, or equivalently |f−1(t)| ≤ 1 for all t ∈ T . It is onto (or epic or surjective) if it maps at least one element of S to

each element of T , i.e. im(f) = T In other words, ∀t ∈ T , ∃s ∈ S with f(s) = t, or equivalently |f−1(t)| ≥ 1 for all t ∈ T .

Exercise Classify x 7→ x2 as a map R → R, R+ → R, R → R+ , R+ → R+

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Proposition 1.2 f : S → T is

aimonic ⇐⇒ ∃` : T → S with l ◦ f = idS (` is a “left inverse” of f)

biepic ⇐⇒ ∃r : T → S with f ◦ r = idT (r is a “right inverse” of f)

Proof (Sketch) ( ai=⇒) For t ∈ im(f), define `(t) to be the (unique) s ∈ f−1(t), and define `(t) arbitrarily for t ∈/ im(f). ( bi=⇒) Define r(t) = any s ∈ f−1(t).

Definition If f is both one-to-one (also written 1-1) and onto, then it is called a bijection, and has an inverse function f−1 : T → S defined by f−1(t) = the unique elt s ∈ S for which f(s) = t.

Note f−1 is characterized by f ◦ f−1 = idT ; f−1 ◦ f = idS .

§2. Equivalence Relations

Definition A partition of a set S is a collection of nonempty disjoint subsets of S whose union is S.

Examples (including Banach-Tarski Paradox)

Definition A relation on S is a subset ∼ of S ×S; write x ∼ y to indicate (x, y) ∈ ∼. Say ∼ is an equivalence relation if it is

1i reflexive : x ∼ x (for all x ∈ S)

2i symmetric : x ∼ y =⇒ y ∼ x

3i transitive : x ∼ y and y ∼ z =⇒ x ∼ z

The equivalence class of x ∈ S is x := {y ∈ S | x ∼ y}.

Proposition 2.1 The collection of equivalence classes of an equivalence` relation on S form a partition of S. Conversely, any partition S = Sαα gives rise to a unique equivalence relation ∼ whose equivalence classes are the Sα’s, namely x ∼ y ⇐⇒ ∃α with x, y ∈ Sα. Proof: exercise

The set of all equivalence classes of an equivalence relation ∼ is called the quotient set of S by ∼

S/ ∼ := {R ⊂ S | R = x for some x ∈ S}

and the map S → S/ ∼, x 7→ x is the natural projection of S onto S/ ∼.

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Examples 1i (the integers mod n) S = Z. For each n ∈ N, define the “congruence mod n” relation ≡n by a ≡n b (also written a ≡ b (mod n)) ⇐⇒ n | (a − b) (i.e. a and b have the same remainder when divided by n). (Check properties 1i, 2i, 3i, e.g. 3i ... ) The equiv class

a = {a + kn | k ∈ Z} = {. . . , a − n, a, a + n, a + 2n, . . . }

is sometimes called the residue class of a (mod n). The quotient set Z/ ≡n, also denoted Zn or Z/nZ, is called the integers mod n. Examples.

2i S = R2 − 0, and x ∼ y ⇐⇒ x = λy for some nonzero real number λ (picture). R2/ ∼, usually denoted RP 1 , is called the real projective line.

§3. Binary Operations

Definition A binary operation on a set S is a function

∗ S × S → S.

Write a∗b for ∗(a, b) (“infix” notation). It is associative if (a∗b)∗c = a∗(b∗c) for all a, b, c ∈ S and commutative if a ∗ b = b ∗ a for all a, b ∈ S.

Examples 1i +, · on S = N, Z, Q, R, C (associative and commutative); − on Z, Q, R, C is neither

2i· of n × n matrices (associative but not commutative)

3i+, · on Zn (“modular arithmetic”):

a + b := a + b

(the LHS is defined by the RHS). Must show this is “well-defined”: Suppose 0 b ≡ b0a ≡ a , , i.e. n|(a − a0), n|(b − b0). Then n|[(a − a0) + (b − b0)] =⇒

n|[(a + b) − (a0 + b0)]. Thus a + b = a0 + b0. Similarly define

a · b := a · b

and show well defined. These operations are associative and commutative. Also · is distributive over +:

a(b + c) = ab + ac.

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Definition A morphism of sets w/ binary operations f : (S, ∗) → (S0 , ∗0) is a function f : S → S0 satisfying

f(a ∗ b) = f(a) ∗0 f(b)

for all a, b ∈ S. Say f is a monomorphism if f is 1-1, and an epimorphism if f is onto. An isomorphism is a morphism which has an inverse which is also a morphism.†

exp xExamples 1i The “exponential map” (R, +) −→ (R, ·), x 7→ e , is a x+y x ymorphism, since e = e e .

det2iMn(R) −→ R is a multiplicative morphism, since det(AB) = det(A) det(B).

†In fact, any morphism f which has an inverse f−1 (or equivalently any bijective morphism) is an isomorphism. You are asked to show this in the homework. This entails showing that f−1(x ∗0 y) = f−1(x) ∗ f−1(y) for any x, y ∈ S0 . To do this, start by noting x = f(a) and y = f(b) for some a, b ∈ S. Now plug in . . .

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II Groups

§1. Basic Concepts: groups, homomorphisms, and group actions

∗ Definition A group is a set G with a binary operation G × G → G, (a, b) 7→ a ∗ b, satisfying (G1) associativity: (a ∗ b) ∗ c = a ∗ (b ∗ c) for all a, b, c ∈ G

(G2) identity: ∃e ∈ G s.t. e ∗ a = a = a ∗ e for all a ∈ G

(G3) inverses: ∀a ∈ G, ∃b ∈ G s.t. a ∗ b = e = b ∗ a. If ∗ is commutative, say G is abelian. |G| is the order of G. If |G| < ∞, say G is finite; otherwise G is infinite. The order of an element a ∈ G is

|a| := min{n ∈ N | |a ∗ ·{z· · ∗ a} = e} n factors

unless this set is empty, in which case |a| = ∞ by convention (e.g. every nonzero integer has infinite order, and |0| = 1).

Remarks 1i Identities and inverses are unique Proof: Suppose e1, e2 are both identities. Then

e1 = e1 ∗ e2 G2 (e2 is an identity) = e2 G2 (e1 is an identity)

Now suppose b1, b2 are both inverses of a. Then

b1 = b1 ∗ e

= b1 ∗ (a ∗ b2) = (b1 ∗ a) ∗ b2

= e ∗ b2

= b2

G2 G3 G1 G3 G2

2iUsually write

· for ∗ and a · b or ab for a ∗ b

1 (or 1G) for e

−1 na for the inverse of a, a for a ∗ · · · ∗ a (n times) when n > 0, and for (a|n|)−1 when n < 0.

If G is abelian, sometimes write + for ∗; a + b, 0, −a, and na for a ∗ b, e, the (additive) inverse of a, and a ∗ · · · ∗ a (n times).

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3iOther structures ai semigroup G1 only (set with an assoc bin op)

bi monoid G1 and G2 only (set with an assoc bin op with identity)

· ci ring abelian group (R, +) with an add’l assoc binary op R × R → R which is distributive over +, i.e. a(b+c) = ab+ac and (a+b)c = ac+bc (e.g. (Z, +, ·)). Say R is commutative if · is, with identity if ∃1.

di field commutative ring (F, +, ·) with 1 =6 0 such that each nonzero elt has a mult inverse (e.g. Q, R, C, and Zn for prime n; HW).

Cancellation Property If a, b, c ∈ G (a group) satisfying ab = ac, then b = c.

Proof (supply reasons) ab = ac =⇒ a−1(ab) = a−1(ac) =⇒ (aa−1)b = (aa−1)c =⇒ 1b = 1c =⇒ b = c.

Familiar examples of groups

1i(Z, +) infinite abelian group. In fact (Z, +, ·) is a commutative ring with 1, but not a field (@ inverses in gen’l)

2iQ, R, C under +, and Q − 0, R − 0, C − 0 under ·

3i(Zn, +) finite abelian group

|Zn| = n Zn = {0, 1, · · · , n − 1}

In fact (Zn, +, ·) is a comm ring with 1. (Example: Z2, mult table)

Application (Linear Diophantine Equations) For a, b, c ∈ Z, find all in-teger solutions to ax + by = c.

Solution First reduce to the case d := gcd(a, b) = 1 (a and b relatively prime): If d 6= 1, then either d - c in which case @ any integer solutions, or d | c in which case we simply divide through by d. So we may assume d = 1. Then working in Zb, the equation becomes a x = c. Since d = 1, the GCD

lemma =⇒ ap+bq = 1 for suitable integers p, q. Thus ap ≡ 1 (mod b), i.e. a has an inverse, namely p, in Zb. Thus the unique solution to the equation in Zb is x = p c =⇒ one solution in Z is x0 = pc, y0 = (c−ax0)/b = (c−apc)/b. To get all the solutions, note that we can change x0 by adding any

multiple nb of b. This will change y0 by subtracting na. Thus the full set of solutions is {(pc + nb, (c − apc)/b − na) | n ∈ Z}.

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Example Solve 6x+10y = 14. Equivalent equation 3x+5y = 7. Working mod 5

3x ≡ 7 =⇒ x ≡ 3−1 · 7 ≡ 2 · 7 ≡ 14 ≡ 4

so solutions are {(x, y) = (4 + 5n, −1 − 3n) | n ∈ Z}.

More examples of groups

1iThe circle group S1 = {z ∈ C | |z| = 1} infinite abelian group (in fact a Lie group: a group which is also a manifold ... locally Euclidean)

n2iThe cyclic group of order n, Cn = {z ∈ C | z = 1}. Note that Cn ⊂ S1 . 2πik/n 7→ k.Also (Cn, ·) is “isomorphic” to (Zn, +) via e

2 23i The Klein 4-group V4 = {1, a, b, c} with a = b2 = c = 1, and the product of any two of a, b, c equals the third. This is not isomorphic to C4

(why?). Every group of order 4 is iso to either C4 or V4 (study mult tables).

4i(products) G, H groups =⇒ G × H is a group under the operation (g, h) · (g0, h0) := (gg0, hh0). (For example, S1 × S1 is the torus (picture), and C2 × C2 = {(1, 1), (1, −1), (−1, 1), (−1, −1)} ∼= V4.) Note ai The product of abelian gps is abelian bi |G × H| = |G||H|.

5i(units) Let (R, ·) be a monoid (e.g. ignore + in any ring R), and set

R• 0 0= {a ∈ R | ∃a 0 ∈ R with aa = a a = 1}

i.e. the “invertible elts” of R. The elements in R• are called units in R.

Claim (R• , ·) is a group (Proof Must first show that · induces an opera-tion on R• , i.e. R• is closed under multiplication: if a, b ∈ R• , i.e. ∃a0, b0 ∈ R

0 0with aa = a a = 1 = bb0 = b0b, then a0, b0 ∈ R• =⇒ (ab)(b0a0) = 1 = (b0a0)(ab) =⇒ ab ∈ R• . Associativity is inherited. 1 ∈ R, since 1 · 1 = 1, and any a ∈ R• has an inverse in R• , namely the a0 from above.)

Examples ai Z• = {+1, −1} = C2

biZ• = {a ∈ Zn | ∃b with ab = 1} = {a ∈ Zn | gcd(a, n) = 1}n def claim

(Proof: a ∈ Z• ⇐⇒ ∃x, y s.t. ax + ny = 1 ⇐⇒ gcd(a, n) = 1)n GCD

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Thus Z• = {a | 0 < a < n, (a, n) = 1}.n

For example Z• = {1, 3, 5, 7} is a gp of order 4 with “multiplication table” 8

· 1 3 5 7 1 1 3 5 7 3 3 1 7 5 5 5 7 1 3 7 7 5 3 1

We know two other gps of order 4: C4 and C2 ×C2. These are not isomorphic to each other: in C2 × C2, all squares = 1 (as in Z•), but not so in C4. Are8 Z• and C2 × C2 isormorphic? Yes: 8

1 ←→ (1, 1) 3 ←→ (1, −1) 5 ←→ (−1, 1) 7 ←→ (−1, −1)

The order |Z• | is called the Euler phi function of n, denoted ϕ(n).n

e1 ekFact Let n = p · · · p where p1, . . . , pk are distinct primes. Then1 k 1 1

aiϕ(n) = n(1 − ) · · · (1 − ). For example: p1 pk

1 1 1 18 ϕ(76) = ϕ(22 · 19) = 76(1 − )(1 − ) = 2219 · · = 36

2 19 2 19

∼biZ• = Z• e1 × · · · × Z•

ek (where =∼ means “isomorphic to”). Furthermore, n p p1 k ∼if n is a power of a prime p, then Z• = Cϕ(n) for odd p and C2 × Cϕ(n)/2 forn p = 2. For example, Z• ∼

76 = C2 × C18. Examples of groups (continued)

6iDihedral groups D2n := {symmetries of a regular n-gon P ⊂ C}

(P has vertices at e2πik/n) A symmetry of P is a rigid motion (i.e. distance preserving function f : C → C) which carries P onto itself (i.e. f(P ) = P );

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each such is either a rotation about 0, or a reflection across a line thru 0. Clearly D2n consists of n rotations and n reflections, so |D2n| = 2n. The operation on D2n is composition. Can write each elt of D2n in terms of r (rotation by 2π/n radians, i.e.

mult by ζn) and s (reflection through the x-axis, i.e. conjugation):

D2n = {1, r, r 2 , . . . , r n−1 , s, sr, . . . , sr n−1}

n 2 −1Note the “relations” r = s = 1 and rs = sr (since ζnz = ζn −1 z). Since

−1any word in the letters r, s, r , s−1 can be reduced using these relations to athe unique form s rb (for some a = 0, 1 and b = 0, . . . , n − 1) we say D2n is

n 2 −1generated by r, s with relations r = s = 1, rs = sr , written

n 2D2n = (r, s | r = s = 1, rs = sr −1).

Remark D4 =∼ V4. Also think about D6 and D8, symmetries of the equilateral triangle and the square, and T12, O24 and I60, symmetries of the (solid) tetrahedron, octahedron (or cube) and icosahedron (or dodeca-hedron).

7iSymmetric Groups Set n = {1, . . . , n}. Then

Sn := {bijections n → n}

is a group under composition. Its elements are called permutations (of n symbols or letters), and (Sn, ◦) is called the symmetric group of degree n. Note that Sn is a finite, nonabelian (for n ≥ 3) group of order n!,

|Sn| = n! (why?)

(What about S1 and S2? Exercise S3 ∼= D6)

More generally, for any set A (possibly infinite), the set SA of bijections A → A is a group under composition, the symmetric group on A.

Notation for elements σ ∈ Sn aitwo-row notation – write the numbers 1, . . . , n in the first row and their images σ(1), . . . , σ(n) in the second.

bi cycle notation (more efficient) – break σ ∈ Sn into disjoint cycles: The k-cycle

(i1 i2 i3 · · · ik)

is the permutation which sends each ij to the next ij+1 in the list, sends ik to i1, and leaves all else fixed (draw circular picture). Note that “cyclic

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permutations” of the list, e.g. (i2 i3 · · · ik i1), represent the same cycle. A 2-cycle is also called a transposition

Obvious fact Every σ ∈ Sn can be written uniquely (up to order of cycles and cyclic permutation in each cycle) as a product of disjoint (i.e. non overlapping) cycles, called its cycle decomposition. (Often suppress 1-cycles in notation)� �

1 2 3 4 5Examples 1i = (1 4)(2 5 3)

4 5 2 1 3� � 1 2 3 4 5

2i = (1 4)(2)(3 5) = (1 4)(3 5) = (3 5)(4 1) 4 2 5 1 3

Exercise List all the elements of S3 and S4.

To multiply (i.e. compose) two permutations, first juxtapose their cycle decompositions. What results is a product of cycles that might not be disjoint. To rewrite this in disjoint cycle form, work from right to left (as with composition of functions) to see where each number 1, . . . , n maps. For example, to compute π = στ where σ = (2 1 4 5 3) and τ = (1 5)(2 3),

first see where 1 maps: we have τ(1) = 5 and σ(5) = 3, and so π(1) = 3. Similarly π(3) = 1 (giving (1 3) as one of the cycles in π), π(2) = 2 (giving the cycle (2)), π(4) = 5 and π(5) = 4 (giving the cycle (4 5)). Thus

(2 1 4 5 3) · (1 5)(2 3) = (1 3)(2)(4 5) = (1 3)(4 5).

Note The order (as an elt of Sn) of any k-cycle is k, and in gen’l the order of a permutation is the least common multiple (lcm) of the orders of the (disjoint) cycles in its cycle decomposition (why?). For example

|(2 1 4 5 3)(6 9)(7 8)| = 10 |(2 1 4 5 3)(1 5)(2 3)| = 2

(not 10 for the latter, since the cycles are not disjoint).

Two final examples

8iQuaternion groups The set of quaternions is

H = {a + bi + cj + dk | a, b, c, d ∈ R}

where i2 = j2 = k2 = −1, ij = k = −ji, jk = i = −kj, ki = j = −ik. Note that H is a group under addition, but not under multiplication. However, it is a ring, in fact almost a field; the only axiom that fails is commutativity of multiplication. Such a structure is called a division ring.

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Exercise The multiplication in H is related to the dot and cross products in R3: Writing a quaternion a + bi + cj + dk as a +(b, c, d) = a + ~v, we have

(a + ~v)(b + w~) = (ab − ~v • w~) + (aw~ + b~v + ~v × w~ ).

The finite quaternion group

Q8 := {±1, ±i, ±j, ±k}

is a nonabelian (multiplicative) subgroup of H• of order 8; it lies in the infinite subgroup

2 + d2S3 := {a + bi + cj + dk ∈ H | a 2 + b2 + c = 1}

called the 3-sphere which arises in topology.

9iMatrix groups R = commutative ring with 1 For each n ∈ N have the group GLn(R) of invertible n × n matrices with entries in R, called the general linear group (over R). This is just the group of units in Mn(R) (the ring of n × n matrices/R).

Note A is invertible iff det(A) is invertible, i.e. in R• (= R − 0 for a field).� � � � � � � � � � � � 1 0 1 1 1 0 0 1 1 1 0 1

Example GL2(Z2) = { , , , , , }0 1 0 1 1 1 1 0 1 0 1 1

n−1Exercise Show that for p prime, |GLn(Zp)| = Q

(pn − pk).k=0

There are many important subgps of GLn(R): SLn(R) = {A | det(A) = 1}. For R = R have O(n) (orthogonal matrices) and SO(n). For R = C have U(n) (unitary matrices) and SU(n). Tricky exercise: Show SU(2) ∼= S3 .

Homomorphisms

Definition Let G, H be groups. A function f : G → H is a group homomorphism (or morphism) if f(xy) = f(x)f(y) for all x, y ∈ G. The image and kernel of f are

im(f) := {f(x) | x ∈ G} = f(G)

ker(f) := {x ∈ G | f(x) = 1} = f−1(1)

For ex the trivial morphism, defined by f(x) = 1 for all x ∈ G, has kernel G and image = {1}. Define mono, epi, and iso-morphisms in usual way (mono

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= 1-1, epi = onto, and iso = both). We say G and H are isomorphic if ∃ and isomorphism G → H. A morphism from a group to itself (i.e. G = H) is called an endomorphism. An automorphism is a bijective endomorphism.

Examples 1i exp : (R, +) → (R• , ·) is a monomorphism.

2idet : GLn(R) → R• is an epimorphism (where it is understood that the operation is multiplication in both groups).

3iFor any abelian group G, the map φ− : G → G, x 7→ x−1 is a homomor-−1 −1 −1 −1phism since φ−(xy) = (xy)−1 = y x = x y = φ−(x)φ−(y). In fact

φ− is an automorphism which is its own inverse! If G is nonabelian, then φ− is not a morphism: for any a, b ∈ G with

−1b−1ab 6= ba, have φ−(ab) = (ab)−1 and φ−(a)φ−(b) = a = (ba)−1 , but (ab)−1 6= (ba)−1 since inverses are unique.

Remarks 0i Any composition of morphisms is a morphism (verify)

1iLet f : G → H be a homomorphism of groups. Then

ai f is onto ⇐⇒ im(f) = H (true for all functions) and f is 1 − 1 ⇐⇒ ker(f) = {1} (HW).

bi f(1) = 1 (proof: f(1) = f(1 · 1) = f(1)f(1); now multiply by f(1)−1) and f(x−1) = (f(x))−1 (HW)

nci |f(x)| divides |x| Proof Set n = |x| and k = |f(x)|. Then x = 1 =⇒ f(x)n = f(xn) = f(1) = 1. Now appeal to:

Order Lemma Let a be an element of order k in a group. If k is finite, then ai = aj ⇐⇒ i ≡ j (mod k). In particular an = 1 ⇐⇒ k|n. If k = ∞

ithen a 6= aj unless i = j.

Proof If k is finite, then for any i and j we can write i − j = qk + r with 0 ≤ k < r. Thus ai ≡ aj (mod k) ⇐⇒ ai−j = 1 ⇐⇒ ar = 1, since aqk+r = (ak)qar = ar . But ar = 1 ⇐⇒ r = 0, since r < k = |a|, and r = 0 ⇐⇒ i ≡ j

i i−j(mod k). The last statement is clear since a = aj ⇐⇒ a = 1.

2iIf f : G → H is an isomorphism of groups, then

ai |G| = |H|

bi G is abelian ⇐⇒ H is abelian

ci |f(x)| = |x| ∀x ∈ G (HW)

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Can use these to prove two groups are not isomorphic. For example:

∼aiCm =6 Cn for m =6 n since they have different orders.

∼biC6 6= S3 since C6 is abelian and S3 is not.

ci Z• ∼= 24 Can also24 6 C8 since C8 has elts of order 8, but Z• doesn’t. count the number of elements of a given order to distinguish groups, e.g.: D24 6=∼ S4 since D24 has 13 elts of order 2 while S4 has 9 (why?).

Open Problem Classify all groups (up to isomorphism) Facts 1i There is only one of any given prime order (prove later)

2i There exist 2 of order 4 (C4, V4), 2 of order 6 (C6, S3), 5 of order 8 (C8, C4 × C2, C2 × C2 × C2, D8, Q8), 2 of order 9 (C9, C3 × C3), 2 of order 10 (C10, D10), 5 of order 12, 14 of order 16 . . . (see page 168 in text)

Group Actions (a central theme in the course)

Definition An action of a group G on a set A is a map G × A → A, (g, a) 7→ g · a satisfying

(A1) g · (h · a) = (gh) · a (A2) 1 · a = a

for all g, h ∈ G, a ∈ A. The associated permutation representation is the function σ : G → SA (the symmetric group on A) defined by

σ(g)(a) = g · a. (∗)

Note that σ is a group homomorphism, and conversely any homomorphism σ

G → SA gives rise to a group action given by (∗) which has σ as its permu-tation representation (exerscise). The kernel of the action is ker(σ) = {g ∈ G | g · a = a for all a ∈ A}.

The action is said to be faithful if it has trivial kernel (=⇒ σ is one-to-one), i.e. g · a = g · b =⇒ a = b.

Examples 1i trivial action g · a = a for all a ∈ A (i.e. σ is the trivial homomorphism). This is not faithful (unless G = 1).

2iThe (faithful) action of D2n on the set of vertices of the n-gon.

3iThe actions of any group G on itself by left or right multiplication: g ·a = ga or ag−1 (verify that these are both actions) or by conjugation

g · a = gag −1

(HW) Note gag−1 is called the conjugate of a by g.

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§2. Subgroups

Definition A subset H of a group G is a subgroup of G, written H < G, if it is a group under the operation induced from G, i.e. (S1) x, y ∈ H =⇒ xy ∈ H (S2) 1 ∈ H (S3) x ∈ H =⇒ x−1 ∈ H

(Note: associativity is automatic)

Examples 1i Every group has the trivial subgroups {1} and G. Any other subgroup will be called proper.

2i For k = 0, 1, 2, . . . , the set kZ = {nk | n ∈ Z} of all multiples of k is a subgroup of Z; there are no others. (Note that + is the operation in Z, so (S2) reads 0 ∈ kZ.) 0Z = {0} and 1Z = Z are trivial, 2Z = evens, etc.

3i{1, r, . . . , rn−1} and {1, s} are subgps of D2n.

4i{1, −1} and {1, i, −1, −i} are subgroups of Q8.

5iCn < S1 < C• .

Subgroup Criterion A subset H of a group G is a subgroup if and only if ai H is nonempty, and bi x, y ∈ H =⇒ xy−1 ∈ H.

Proof (=⇒) 1 ∈ H by (S2) so H =6 ∅. If x, y ∈ H, then y−1 ∈ H by (S3) so xy−1 ∈ H by (S1). Thus bi holds.

−1(⇐=) ∃x0 ∈ H by ai, so 1 = x0x ∈ H by bi =⇒ (S2). Now 0 −1x ∈ H =⇒ x = 1·x−1 ∈ H by bi=⇒ (S3). Finally x, y ∈ H =⇒ y−1 ∈ H

by (S3) =⇒ xy = x(y−1)−1 ∈ H by bi, so (S1) holds. Remark If H is finite, bi can be replaced by (S1), since x ∈ H =⇒

n −1 n−1 ∈ H.x = 1 for some n (since G is finite) =⇒ 1 = xn ∈ H and x = x

More examples of subgroups

6iLet G be a group. Then any subgp of a subgp of G is a subgp of G, and any intersection of subgps of G is a subgp of G (why?).

f7iFor any homomorphism G → H of groups,

ker(f) < G and im(f) < H

(check this by the def’n of subgp or using the subgp criterion)

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8i The center Z(G) := {x ∈ G | xg = gx for all g ∈ G} of a group G is a subp of G. Proof: For x, y ∈ Z(G) and g ∈ G, we have xyg = xgy =

−1 −1gxy =⇒ xy ∈ Z(G) (proving S1) and xg = g x =⇒ (taking inverses) −1 −1x g = gx =⇒ x−1 ∈ Z(G) (proving S3). For (S2), note that 1 ∈ Z(G) since 1g = g = g1 for all g. (Exercise Give an alternative proof using the subgp criterion and the observation that xg = gx ⇐⇒ x = gxg−1)

More generally, for any subset A ⊂ G, define the centralizer of A in G to be

CG(A) := {x ∈ G | xa = ax for all a ∈ A}

(so Z(G) = CG(G)), and the normalizer of A in G to be

NG(A) := {x ∈ G | xA = Ax}

where xA = {xa | a ∈ A} and Ax = {ax | a ∈ A}. All are subgroups of G (general proof given in 9i below).

Example Z(Q8) = {1, −1} (verify). For A = {1, −1, i, −i} < Q8, have ji 6= ij, ki =6 ik, etc. and so CQ8 (A) = A. However jA = {j, −j, −k, k} is the same set as Aj = {j, −j, k, −k}, . . . and so NQ8 (A) = Q8.

Remark In general, for any subsets A1, . . . , An of a group G, define

A1 · · · An = {a1 · · · an | ai ∈ Ai for each i = 1, . . . n}.

xA and Ax are special cases of this, as is xAx−1 = {xax−1 | a ∈ A}. Note that the conditions xa = ax and xA = Ax defining CG(a) and NG(A) can be rewritten as xax−1 = a and xAx−1 = A.

9iIf a group G acts on a set A and a ∈ A, then the stabilizer of a is

Ga = {x ∈ G | x · a = a}

This is a subgroup of G for each a. (Proof: 1 ∈ Ga by (A2), so Ga 6= ∅. −1 −1 −1Thus for any x ∈ Ga, x · a = x · (x · a) = (x x) · a = 1 · a = a so

x−1 ∈ Ga. If x, y ∈ Ga then (xy) · a = x · (y · a) = x · a = a so xy ∈ Ga.)

Examples ai The stabilizer of 1 ∈ C under the usual action of the dihedral group D2n is {1, s}.

bi Normalizers are examples of stabilizers for a suitable action: Let G act by conjugation on the set of all subsets of G,

x · A = xAx−1 ( = {xax −1 | a ∈ A} ).

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Then for any A ⊂ G,

NG(A) = {x ∈ G | xA = Ax} = {x ∈ G | xAx−1 = A} = GA.

Similarly centralizers can viewed in terms of actions: For A ⊂ G, the nor-malizer NG(A) acts on A by conjugation, and CG(A) is the kernel of this action (do you see why?). Thus the fact that normalizers and centralizers are subgroups can be

deduced from the fact that stabilizers and kernels of homomorphisms are.

Normal subgroups

Definition A subgroup H < G is called a normal subgroup (written H C G) if

h ∈ H, g ∈ G =⇒ ghg−1 ∈ H

(or equivalently NG(H) = G). Thus a subgroup of G is normal iff it is “closed under conjugation” by any elt in G. For example {1, −1, i, −i} C Q8, as shown above. In general

1i Z(G) C G (why?)

2i ker(f) C G for any morphism f : G → H (why?).

13i Any H < G with |H| = |G| (say H is of index 2 in G) is normal in G2

Proof of 3i For any x 6∈ H, the set xH is disjoint from H (if xh = h0 for some h, h0 ∈ H then x = h−1h0 ∈ H =⇒⇐=). So xH = G − H. Similarly Hx = G − H. Thus xH = Hx =⇒ xHx−1 = H =⇒ H C G.

Warning: K C H C G =6 ⇒ K C G. ∃ examples with |G| = 8 (HW).

Cyclic groups and subgroups

Definition For any element x in a group G, denote by hxi the set of all powers of x (or multiples of x if G is additive),

khxi = {x | k ∈ Z} (w/ repetitions deleted)

This is a subgroup of G (verify) called the cyclic subgroup generated by x. If

G = hxi for some x ∈ G

we say G is a cyclic group, and any such x is called a generator of G (there may be many). Note: if |G| = n, then G cyclic ⇐⇒ G has an elt of order n.

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Examples 1i Z under + (or its mult analogue C∞ = {tk | k ∈ Z}, where t is an indeterminant) is cyclic with generator 1 (or t). Alternatively, −1 (resp. t−1) is a generator.

2i 2πi/n (or tkCn is cyclic with generator t = e for any k rel prime to n)

k k3iD2n is not cyclic for n > 1: hrki ⊂ hri =6 D2n and hr si = {1, r s} 6= D2n

Remark The order n of any element x in a group is equal to the order of the cyclic subgroup it generates: |x| = |hxi|. (Clear if n = ∞. If n < ∞ then xi = xj ⇐⇒ i ≡n j (by the order lemma) so hxi consists of the n distinct elements 1, x, . . . , xn−1.)

Classification Theorem for Cyclic Groups Every cyclic group C = hxi is isomorphic to Cn for some n = 1, 2, · · · , ∞. Thus there is up to isomorphism exactly one cyclic group of each finite order and one infinite cyclic group.

Proof If |C| = n, then the function C → Cn mapping xk to tk (for each k ∈ Z) is a well defined isomorphism (why?).

Classification Theorem for Subgroups of Cyclic Groups Let C = (x) be a cyclic group. Then

ai Every subgroup H of C is cyclic. In particular any nontrivial H = hxmi, where m is the smallest positive integer for which xm ∈ H.

bi If C is finite of order n, then it has a subgroup of order k ⇐⇒ k divides n. In fact there is only one such subgroup for each divisor k.

Proof ai HW bi (=⇒) If H < C with |H| = k, then H = hxmi, for m as in ai. Set d =

d am+bn m)agcd(m, n). Then x = x (for suitable a, b) = (x ∈ H =⇒ m = d, n/ki, so H iswhich divides n =⇒ k = n/m divides n (and in fact H = hx

the only subgroup of order k). (⇐=) k|n =⇒ H = hxn/ki has order k.

This theorem gives a complete picture of the “lattice” of subgroups of any cyclic group (draw pictures, cf. §2.5 in the text).

Remarks 1i ∃ groups with non-planar lattices, e.g. D16

2i ∃ pairs of nonisomorphic groups with the same subgroup lattice, e.g. 8 2C2 ×C8 and the “modular group” of order 16 = (r, s | r = 1 = s , rs = sr5).

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Question Does bi generalize to other finite groups G?

Answer =⇒ does (Lagranges’s theorem, below), but ⇐= does not. For example the group T12 of symmetries of the tetrahedron has order 12 but has no subgroup of order 6. To see this, note that T12 consists of eight 2π/3-rotations (about lines through the vertices), three π-rotations (about lines joining midpoints of opposite sides), and the identity element. Now suppose that H < T12 with |H| = 6. Then it must contain at least one 2π/3-rotation r. But then it must contain all of them (since they are all conjugate to r or −1r , and H C G as shown above) and this is clearly impossible.

Lagrange’s theorem The order of any subgp of a finite gp G divides |G|

Definition If H is a subgroup of a group G, then any subset of G of the form

xH = {xh | h ∈ h}

for x ∈ G, is called a left coset of H in G. Similarly define the right cosets Hx ⊂ G. The left (resp. right) cosets of H partition G into equal sized subsets:

Coset Lemma ai |xH| = |H| bi xH ∩ yH 6= ∅ =⇒ xH = yH, and ci ∪x∈GxH = G (and similarly for right cosets)

Proof ai The map H → xH, h 7→ xh is a bijection, by cancellation bi hyp =⇒ xi = yj for some i, j ∈ H, so xh = xii−1h = yji−1h ∈ H for any h ∈ H =⇒ xH ⊂ yH. Similarly yH ⊂ xH, so xH = yH.ic x ∈ xH.

The number of left (or right) cosets of H in G is called the index of H in G, denoted |G : H|. The coset lemma shows that |G| = |G : H||H|, which proves Lagrange’s Theorem.

Remarks 1i The set of all left cosets is (sometimes) denoted by G/H, so |G/H| = |G : H| = |G|/|H|. Similarly for the set H\G of right cosets.

−12i(coset recognition: HW) x, y ∈ G lie in the same coset of H ⇐⇒ x y ∈ H ⇐⇒ y = xh for some h ∈ H (and similarly for right cosets)

Example of coset decomp: The subgp H = {1, (1 2)} of S3 has cosets H, (1 3)H = {(1 3), (1 2 3)} and (2 3)H = {(2 3), (1 3 2)}.

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Applications of Lagrange’s Theorem

1iThe order of any element x in a finite group G divides |G|.

2iAny group of prime order p is cyclic (and so there’s only one).

3i(Euler’s Theorem) If a, n ∈ N are relatively prime, then

φ(n) ≡ 1a (mod n)

where φ(n) = |Z• |. Special case (Fermat’s Little Theorem): If p is prime and n p - a, then ap−1 ≡ 1 (mod p) (since φ(p) = p − 1). For example 34 = 81 ≡ 1 (mod 5), 36 = 729 ≡ 1 (mod 7), etc.

Proofs 1i |x| = |hxi|, which divides |G| by Lagrange.

2iAny x 6= 1 in G has order dividing p (by 1i) and thus equal to p, since p is prime; so G = (x).

φ(n)3i(a, n) = 1 =⇒ a ∈ Z• (by the GCD Lemma) =⇒ a = 1 (by 1i, sincen Z• has order φ(n)), i.e. aφ(n) ≡ 1 (mod n).n

§3. Products and Finite Abelian Groups

Recall that the product of two groups H and K is

H × K = {(h, k) | h ∈ H, k ∈ K}

with componentwise multiplication (h, k)(h0, k0) = (hh0, kk0). It turns out that every finite abelian group is a product of cyclic groups. More precisely,

Fundamental Theorem of Finite Abelian Groups (Primary Form) Every finite abelian gp is isomorphic to a product of cyclic groups of prime power order. Moreover, the number of factors and their orders are uniquely deter-mined by the group.

This leads to an algorithm for finding all abelian groups of order n: k• If n = p , a pure prime power, then there is (up to isomorphism) exactly

one abelian group of order n for each partition of k (a sequence k1 ≥ · · · ≥ ks

of natural numbers such that k = k1 + · · · + ks), namely

C k1 × · · · × Cpkn . p

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For example, there are three abelian groups of order 125 = 53

C125 C25 × C5 C5 × C5 × C5

corresp to the three partitions 3, 2 + 1 and 1 + 1 + 1 of 3.

k1 k2• For general n = p p · · · , there is one abelian group for each list of1 2 partitions of k1, k2, . . . .

Exercise How many abelian groups are there of order 1500 = 22 · 3 · 53? (Answer: 6 = 2 · 1 · 3.) List them. (C4 × C3 × C125, C2 × C2 × C3 × C125, C4 × C3 × C25 × C5, C2 × C2 × C3 × C25 × C5, C4 × C3 × C5 × C5 × C5, C2 × C2 × C3 × C5 × C5 × C5).

There is another standard form for finite abelian groups: Using the fact (from homework) that

∼Cm × Cn = Cmn

if m and n are relatively prime, can start with the primary form, then group the largest factors associated with each prime in the primary form, then the next largest, etc. to get a unique form for any finite abelian group

G ∼= Cn1 × Cn2 × · · ·

where n = n1·n2 · · · and ni is divisible ni+1 for each i; the numbers n1, n2, . . . are called the invariant factors of G. For example

∼C4 × C2 × C3 × C5 × C5 = C60 × C10

has invariant factors 60, 10. The group C8 × C2 × C28 × C25 × C14 has inv factors 1400, 28, 2, 2 (exercise). In summary

Fundamental Theorem of Finite Abelian Groups (Invariant Form) Any finite abelian gp G is isomorphic to a product Cn1 × Cn2 × · · · × Cnk where ni is divisible by ni+1 for i = 1, . . . , k − 1. The numbers n1, . . . , nk, called the invariant factors of G, are unique.

We’ll prove the first statement in the fundamental theorem (primary form). Need

Product Recognition Theorem (PRT) If H and K are normal subgroups of a group G satisfying H ∩ K = 1 and HK = G, then G =∼ H × K.

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Proof Consider the function f : H × K → G given by f(h, k) = hk, which is onto since HK = G. We claim that f is a homomorphism. First note that for any h ∈ H and

k ∈ K, hkh−1k−1 ∈ H ∩ K (it can be written as h(kh−1k−1) ∈ H since H C G, or as (hkh−1)k−1 ∈ K since K C G). Since H ∩ K = 1, it follows that hkh−1k−1 = 1, i.e. hk = kh, so the elts of H commute with the elts of K. Now f((h, k)(h0, k0)) = f(hh0, kk0) = hh0kk0 = hkh0k0 = f(h, k)f(h0, k0). Finally observe that ker(f) = 1 (since f(h, k) = 1 =⇒ hk = 1 =⇒ h =

k−1 ∈ H ∩ K = 1, i.e. (h, k) = (1, 1)) so f is 1-1.

Now let G be a finite abelian group. We wish to show that G is a product of cyclic groups of prime power orders. We split off one prime at a time:

kSuppose |G| = p q, where p is prime and p - q. Set

p qP = {x ∈ G | x k = 1} and Q = {x ∈ G | x = 1}.

Then P and Q are subgroups (since G is abelian) with P ∩ Q = 1 (since p - q) and PQ = G: ∃ a, b with aq + bpk = 1, so for any x ∈ G

aq+bpk aq bpk x = x = x x ∈ PQ

(do you see why?) Thus G ∼= P × Q by the PRT. It remains (by induction on n) to prove that P is a product of cyclic groups. This is the content of the following theorem:

Theorem If P is an abelian p-group for some prime p (meaning that each element in P has order a power of p) then P is a product of cyclic groups.

Proof Let H be the cyclic subgroup of P generated by an element h of maximal order (say ps) in P , and K be a largest possible subgroup of P for which H ∩ K = 1. If the subgroup HK = P , then P =∼ H × K by the PRT, and the theorem follows by induction on |P |. So assume HK 6= P . We will show that this leads to a contradiction.

Claim ∃ x ∈ P such that x ∈/ HK but xp ∈ K.

Pf of claim First note that ∃ y 6∈ HK with yp ∈ HK. (Indeed, for any m m−1p pz 6∈ HK, choose the smallest m such that z ∈ HK, and set y = z .)

pNow y = hnk for some n ∈ Z, k ∈ K. By the maximality of |h| we have s s−1 s−1 s−1p = hnpy = (hnk)p kp = 1.

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h−n/pThus hnps−1 = 1, since H ∩ K = 1, and so p|n. Set x = y. Clearly

px = k ∈ K, but x 6∈ HK since y 6∈ HK. This completes the proof of the claim.

Now let K 0 be the subgroup generated by x and K,

K 0 = {x nk | n ∈ Z, k ∈ K}.

nObserve that H ∩ K 0 = 1, since xnk = h ∈ H =⇒ x = hk−1 ∈ HK =⇒ np|n =⇒ x ∈ K =⇒ h ∈ K =⇒ h = 1. But K 0 % K, which contradicts the

maximality of K.

§4. Cayley’s Theorem and the Symmetric Group

Cayley’s Theorem Every group G is isomorphic to a subgroup of the symmetric group SG.

Proof Let G act on itself by left multiplication. Then the associated permutation homomorphism λ : G → SG (λ(g) = left mult by g) is 1-1 (by the cancellation property in G) and so G ∼= im(λ) < SG.

In particular, every finite group of order n is isomorphic to a subgroup of Sn. One especially important subgroup of Sn is the alternating group An

consisting of all even permutations (defined below). Note that every permutation σ ∈ Sn can be written as a product of

transpositions, since any cycle can:

(i1 · · · ik) = (i1 i2)(i2 i3) · · · (ik−1 ik)

(e.g. (i j k l) = (i j)(j k)(k l)). This decomposition is not unique, e.g. 1 = (1 2)(1 2) = (1 2)(3 4)(1 2)(3 4), but the parity (even or odd) of the number of transpositions is always the same:

Definition A permutation is even if it can be written as a product of an even number of transpositions, and is odd if it can be written as a product of an odd number of transpositions.

In particular odd-length cycles are even and even-length cycles are odd (by the formula displayed above). Also even·even = odd·odd = even, and even·odd = odd·even = odd.

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Remarkable Fact No permutation is both even and odd.

There are many proofs (cf. pages 109–111 in the text, or the enlightening dance floor proof by Ty Cunningham in Math. Magazine 43 (1970) 154-5). Here is one: Consider the function

c : Sn → N

assigning to σ ∈ Sn the number of cycles in the disjoint cycle decomp of σ (counting the 1-cycles). For example c(k-cycle) = n − k + 1. Readily verify

c(στ ) ≡ c(σ) + 1 (mod 2)

for any transposition τ = (i j). Indeed if i and j lie in the same cycle in σ then that cycle splits into two in στ , and if they lie in distinct cycles in σ then those cycles are joined into one in στ . The remaining cycles of σ and στ coincide. It follows that if σ is a product of m transpositions then c(σ) ≡ c(1) + m ≡ n + m, and so the parity of m is determined by σ.

Thus there is a well-defined homomorphism

sgn : Sn → C2 = {±1}

sending even permutations to +1 and odd ones to −1. The kernel is the set An of all even permutations, known as the alternating group of degree n. It is a normal subgroup of index 2, i.e. |An| = n!/2 (do you see why?).

Remarks 1i The parity of a permutation σ is equal to the parity of

ai(algebraic) the number of even-length cycles in any cycle decomposition of σ; the number of odd-length cycles is irrelevant. Thus even permutations are those with even number of even-length cycles: (· · ·), (· ·)(· ·), (· · · · ·), (· · ·)(· · ·), (· · · ·)(· ·), (· ·)(· ·)(· ·), ... This helps enumerate the elements in An: A3 = {1, (1 2 3), (3 2 1)}; A4 = {1, (1 2 3), · · · , (1 2)(3 4), · · · }, etc.

bi(geometric) the number of intersection points in any “picture” of σ

2iConjugation in Sn: To compute στσ−1 , write τ as a product (i j · · · ) of cycles; then

σ(i j · · · )σ−1 = (σ(i) σ(j) · · · )

i.e. replace each number in τ by its image under σ. This is easy to see if τ is a single cycle, and generalizes using the trick στ1τ2 · · · τrσ

−1 = (στ1σ

−1)(στ2σ−1) · · · (στrσ

−1). It follows that if two elements in Sn are conjugate, then they have the same cycle structure. The converse is also

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true; indeed it is an easy to determine all σ for which στσ−1 = τ 0 for any τ and τ 0 which have the same cycle structure. Conjugation in An is more complicated; one must check whether any of

the possible conjugating permutations is even. For example τ = (2 3 4) and τ 0 = (4 3 2) are conjugate in S4 (why?) but not in A4 (since στσ−1 = τ 0 =⇒ σ maps the ordered triple (2, 3, 4) to either (4, 3, 2), (3, 2, 4) or (2, 4, 3), which forces σ = (2 4), (2 3) or (3 4), all of which are odd). This is explored further in the HW.

∼ ∼3i A1 and A2 are trivial groups; A3 = C3; A4 ∼= T12, A5 = I60.

4iAn is generated by 3-cycles. Indeed each element of An can be written as a product of an even number of transpositions, and (i j)(j k) = (i j k), (i j)(k `) = (i j k)(j k `).

5i Recall that An C Sn. In fact if n ≥ 5, then An is the only proper normal subgroup of Sn. Proof: Let 1 6= K C Sn. Then any σ =6 1 in K has a disjoint cycle

decomposition of the form (i j · · · ), and so does not commute with the transposition τ = (j k), for any chosen k 6= i, j. Indeed στ (i) = j whereas τσ(i) = k. Now consider the “commutator” [σ, τ ] := στσ−1τ−1 6= 1. Clearly [σ, τ ] ∈ K (since [σ, τ ] = σ(τσ−1τ−1) and K is normal) and [σ, τ ] is a product (στσ−1)(τ−1) of two distinct transpositions (by 2i). If these transpositions overlap, then [σ, τ ] is a 3-cycle =⇒ K contains all 3-cycles (since it is normal) =⇒ K ⊂ An (by 4i) =⇒ K = An or Sn. If they don’t overlap, then K contains all products of pairs of disjoint transpositions (since it is normal), and in particular (1 2)(3 4) and (3 4)(2 5), whose product is the 3-cycle (1 2)(2 5) = (1 2 5) =⇒ K contains all 3-cycles and so K = An or Sn.

Note We have repeatedly used the observation that a normal subgroup that contains an element h must contain all conjugates of h. In fact H is normal in G if and only if it is a union of conjugacy classes in G. (The conjugacy class of x ∈ G is the set of all elts in G that are conjugate to g.)

In general, one can understand a lot about the structure of a group from a knowledge of its normal subgroups. A group which has no proper normal subgroups is called a simple group; these are the building blocks of all finite groups. (More on this later)

Examples 1iAny group of prime order is simple (indeed such a group has no proper subgroups whatsoever by Lagrange’s Theorem).

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2iAbel’s Theorem An is simple for all n ≥ 5.

Proof Suppose ∃ K C An, but K =6 An or 1. Then An is the normalizer of K in Sn, by remark 5i above, and so K has exactly two “conjugates” in Sn; i.e. setting H = (1 2)K(1 2), we have τKτ−1 = K or H according to whether τ is even or odd (check this). But then H ∩ K C Sn and so H ∩ K = 1, by 5i again. It follows that the elements of H commute with the elements of K (note H C An). But if σ 6= 1 ∈ K, then ∃ odd τ such that σ and τστ−1 do not commute (for example, if σ = (i j k · · · ) or (i j)(k `) · · · , take τ = (k m) for any m 6= i, j, k, and consider the image of j or `, respectively), which is a contradiction.

§5. Quotient Groups

Recall that normal subgroups can be characterized in a variety of ways.

Normality Lemma A subgroup H of G is normal in G if and only if any of the following conditions holds:

aiH is a union of conj classes in G bi xhx−1 ∈ H for all h ∈ H, x ∈ G

cixHx−1 ⊂ H for all x ∈ G di xHx−1 = H for all x ∈ G

eixH = Hx for all x ∈ G fi xHyH = xyH for all x, y ∈ G

Proof The equivalence of ai– ei was noted previously, so it suffices to show ei =⇒ fi =⇒ ci: Assuming ei, have xHyH = xyHH = xyH (HH ⊂ H since H < G, and HH ⊃ H since 1 ∈ H). Assuming fi, have xHx−1H = H =⇒ xHx−1 ⊂ H since 1 ∈ H.

Using fi, can make the set G/H of left cosets of H in G into a group, called the quotient group of G by H, provided H is normal in G:

Theorem If H C G, then G/H is a group under the operation defined by xH · yH = xyH. Furthermore the map

p : G → G/H

defined by p(x) = xH, called the natural projection of G onto G/H, is an epimorphism.

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Proof The operation is well defined (by the lemma), associative, with identity element 1H = H and (xH)−1 = x−1H (verify). Clearly p is a homomorphism, since p(xy) = xyH = xHyH = p(x)p(y), and is onto since every coset is the image of any element in it.

Remarks 1iThe product of two cosets H1, H2 in G/H can be described in words as follows: Choose one element from each. Then H1H2 is the coset containing the product of these elements. The lemma shows that this is “well-defined”, i.e. independent of which elements you choose.

2i Be very careful when G is an “additive” group, i.e. the operation is +. Then cosets of H < G are of the form x + H = {x + h | h ∈ H}, and the operation in G/H is addition: (x+H)+(y +H) = (x+y)+H. For example, if H = nZ = {multiples of n} < Z, then Z/nZ = {k +nZ | k = 0, . . . , n−1}. Note: If G0 is another additive group, then the product G × G0 will be

often be called the direct sum, written

G ⊕ G0 = {(g, g 0) | g ∈ G, g 0 ∈ G0}

with the operation (g, g0) + (h, h0) = (g + g0, h + h0). In other words G ⊕ G0

is the same group as G × G0 , but written additively (cf. example 2ibelow). 3iAny quotient group G/H of an abelian group G is abelian, since xHyH = xyH = yxH = yHxH, for any x, y ∈ G.

Examples 1i Z/nZ =∼ Zn, via the isomorphism k + nZ 7→ k.

2i Let H := h4i = {0, 4, 8} C Z12. Then H has order 3 =⇒ Z12/H = {H, 1 + H, 2 + H, 3 + H} has order 4 =⇒ Z12/H ∼= C4 or V4. Since 1 + H has order 4 (why?) we have in fact Z12/H ∼= C4.

3iLet S := h(0, 1)i C G := Z2 ⊕Z4. Note that |S| = 4 (since (0, 1) has order 4) so G/S has order |G/S| = |G|/|S| = 4/2 = 2, and therefore G/S ∼= Z2. G also has three subgroups of order two, H = h(1, 0)i, J = h(1, 2)i and K = h(0, 2)i, with G/H ∼= H/J ∼= Z4 and G/K ∼= Z2 ⊕ Z2; verify this by computing the orders of elements in the quotient, e.g. (0, 1) + H has order 4 in G/H.

4i Let H be the cyclic subgroup of S3 generated by the 3-cycle (1 2 3), H = h(1 2 3)i = {1, (1 2 3), (1 3 2)} C S3. Then H has two cosets, H and H 0 = (1 2)H = {(1 2), (2 3), (1 3)}, so S3/H = {H, H 0} =∼ C2.

5iZ(D8) = {1, r2} C D8, and D8/Z(D8) =∼ C2 ×C2 (compute orders again).

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�

�

Most of the important properties of quotient groups follow from the

Universal Property of Quotient Groups Let K C G and p : G → G/K be the natural projection. Then for any homomorphism f : G → H whose kernel contains K, there exists a unique homomorphism g : G/K → H such that f = g ◦ p, i.e. the following diagram commutes

pG −→ G/ ker(f)

f & ↓ g

H

In particular g(xK) = f(x), i.e. g(coset) = f(any element in the coset). Also ker(g) = ker(f)/K and im(g) = im(f).

Proof First check that g is well defined (and thus unique by the com-−1mutativity of the diagram): if x, y lie in the same coset, then x y ∈ K.

−1But K ⊂ ker(f), by hypothesis, so f(x y) = 1 = f(x)−1f(y) =⇒ f(x) = f(y). It is now straightforward to show g is a homomorphism (g(xKyK) = g(xyK) = xy = g(xK)g(yK)) with ker(g) = {xK | f(x) = 1} = ker(f)/K and im(g) = {f(x) | x ∈ G} = im(f).

The UPQG can be used to construct morphisms from quotient groups. For example, you can show that Z2 ⊕ Z4/h(0, 2)i is isomorphic to Z2 ⊕ Z2

(see example 2i above) by constructing an explicit iso. One such arises from the UPQG via the map Z2 ⊕Z4 → Z2 ⊕Z2, (a, b) 7→ (a, b) (check this). It has the following important consequences:

The Isomorphism Theorems

1i If f : G → H is a homomorphism, then G/ ker(f) ∼= im(f).

2i If H < G and K C G, then H ∩ K C H and HK/K ∼= H/ H ∩ K.

HK � �

H K � �

H ∩ K

3iIf H, K C G with H ⊂ K, then K/H C G/H and G/K ∼= (G/H)/(K/H).

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Proof Apply the Universal Property to

1i G −→ G/ ker(f)

f & ↓

im(f) ⊂ H

2i(note that HK is a subgroup of G)

H −→ H/ H ∩ K

incl ↓ ↓

HK −→ HK/K

3i G −→ G/H

& ↓ ← now apply 1i to this

G/K

The Correspondence Theorem Let f : G → H be a homomorphism of groups, G be the set of all subgroups of G containing K = ker(f) and H be the set of all subgroups of H contained in I = im(f). Then the map

f : G → H , S 7→ f(S)

is a bijection which respects containment, indices, normality and quotients. In other words,

(a) S < T ⇐⇒ f(S) < f(T ), in which case |T : S| = |f(T ) : f(S)|

(b) S C T ⇐⇒ f(S) C f(T ), in which case T/S ∼= f(T )/f(S)

Proof The inverse of f is the “preimage” map f−1 : H → G sending each subgroup A ∈ H to its full preimage f−1(A) ∈ G. Indeed f−1f(S) = S: The inclusion ⊃ holds in general (s ∈ S =⇒

f(s) ∈ f(S), i.e. s ∈ f−1f(S)) and ⊂ holds since S ⊃ K (if x ∈ f−1f(S), i.e. f(x) ∈ f(S) so f(x) = f(s) for some s ∈ S, then f(xs−1) = 1, i.e. xs−1 ∈ K =⇒ x ∈ Ks ⊂ S). Similarly ff−1(A) = A: ⊂ holds in gen’l, and ⊃ since A ⊂ I. Thus f is a bijection. The rest is straightforward, and is left as an exercise for the reader; the

UPQG is used to construct the isomorphism of quotient groups.

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(R) detExamples 1i applied to GLn → R• gives GLn(R)/SLn(R) ∼= R• .

2iapplied to M, N C G for which M ∩ N = 1 and MN = G gives isomor-phisms

G/M ∼= N and G/N ∼= M.

Note: if M and N are simple groups which are “maximal” proper normal subgroups of G (i.e. there are no larger such subgroups) then the conditions M ∩ N = 1 and MN = G are automatic.

Application : Classification of Finite Groups

Definition A composition series for a finite group G is a sequence of subgroups

G = G0 B G1 B G2 B · · · B Gn−1 B Gn = 1

each a maximal proper normal subgroup of the preceeding.

The quotients Gi/Gi+1 are simple, by the correspondence theorem, and G can be viewed as being built up from these composition factors, which are unique up to order by the “Jordan Holder Theorem” (proved using the isomorphism theorems, cf. the special case in 2i above, where n = 1). Unfortunately ∃ distinct groups with identical composition factors, so

the classification of finite groups is a two-step program: the Holder Program (late 19th century):

1i Classify all simple finite groups

2i Find all ways to “put simple groups together” to form other groups

The first step was recently achieved (∼1980): 18 infinite families

1) Cp (p prime)

2) An (n ≥ 5)

3) PSLn(F ) (F a finite field, n ≥ 2) . . . etc.

and 26 “sporadic” (exceptional) simple groups. The smallest sporadic group, of order 7920, was discovered by Mattieu in 1861. The largest, of order

808, 017, 424, 794, 512, 875, 886, 459, 904, 961, 710, 757, 005, 754, 268, 000, 000, 000

was constructed by Fischer-Griess in 1981, the “monster” group.

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§6. Sylow Theory

Throughout this section

G will denote a finite group of order n.

Recall Lagrange’s Theorem: H < G, |H| = k =⇒ k|n. The converse fails: k|n =6 ⇒ ∃H < G, |H| = k. For example

A4 (∼= T12) has no subgroup of order 6

as proved above. (Another pf using quotient groups: Sps H < A4, |H| = 6. Then H C A4, so A4/H ∼= C2 =⇒ H contains all eight 3-cycles τ in A4, since τH = (τH)3 = τ 3H = H =⇒ τ ∈ H. This contradicts |H| = 6.) Partial converses hold:

Cauchy’s Theorem If p is prime and p|n, then ∃H < G with |H| = p.

More generally

kSylow Theorem I If p is prime and pk|n, then ∃H < G with |H| = p .

Definition Let p be a prime. Any group P of order pα is called a p-group, αand if P < G then P is called a p-subgroup of G. If in addition p is the

largest power of p that divides |G|, that is

α n = p m and p - m

then P is called a Sylow p-subgroup of G. They always exist by Sylow I.

For example If n = 500 = 2253 , then G has subgroups of orders 2, 4, 5, 25 and 125. The ones of order 4 are the Sylow-2 subgroups, and those of order 125 are the Sylow-5 subgroups.

Sylow Theorem II (Conjugacy) Any two Sylow p-sugroups P, Q of G are conjugate, i.e. ∃x ∈ G such that Q = xP x−1 .

Sylow Theorem III (Counting) The total number np = np(G) of Sylow p-subgroups of G satisfies ai np|m and bi np ≡ 1 (mod p).

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Example If G has order 12 = 223, then n3|4 and n3 ≡ 1 (mod 3), so n3 = 1 or 4. Either case may arise: n3(C12) = 1 and n3(A4) = 4 (the four Sylow 3-subgroups of A4 are h(123)i, h(124)i, h(134)i and h(234)i, which are all conjugate). Similarly n2 = 1 or 3, e.g. n2(C12) = n2(A4) = 1 while n2(D12) = 3 (can you find the three Sylow 2-subgps of D12?).

Application (non-simplicity results)

Observe that if a finite group G has only one Sylow p-subgroup P , for some prime divisor p of |G|, then P C G (since any xP x−1 is also a Sylow p-subgroup of G). Thus

np(G) = 1 =⇒ G is not simple

if G is not a p-group. (It can also be shown that the only p-group that is simple is Cp, see below). It’s often not hard to show some np = 1:

1i|G| = pq for distinct primes p and q =⇒ G is not simple.

Proof: We may assume p > q, and then np|q and np ≡ 1 (mod p), by Sylow III, which forces np = 1.†

For example no groups of order 6, 10, 14, 15, 21, 22, 26, 33, . . . are simple.

2i|G| = 30 = 2 · 3 · 5 =⇒ G is not simple.

Proof: The possibilities (using Sylow III) are n2 = 1, 3, 5, 15, n3 = 1, 10 and n5 = 1, 6. Thus if G is simple, then n2 ≥ 3, n3 = 10 and n5 = 6. Now observe that if P and Q are distinct Sylow subgroups of G, then

P ∩ Q = 1. This is true in general if P and Q are associated with distinct primes, since P ∩ Q is a subgroup of both P and Q, and so |P ∩ Q| divides both |P | and |Q| =⇒ |P ∩ Q| = 1 (since |P | and |Q| are relatively prime). If P and Q are associated with the same prime p and are of prime order (this is the case for n = 30 since 30 is square free), then P ∩ Q must be trivial since it is a proper subgroup of P and |P | = p. (Note that two Sylow p-subgroups that are not of prime order may overlap nontrivially!) Finally, count the nonidentity elements in G, using the fact that the

Sylow subgroups don’t overlap: |G| > 6 · 4+2 · 10 + 1 · 3 = 47 =⇒⇐=. Thus G is not simple.

†In fact, Sylow theory gives more information. For example, if q - (p − 1), then G is in fact cyclic! Indeed Sylow III then gives nq = 1 as well, so G has a unique Sylow p-subgroup P ∼= Cp and a unique Sylow q-subroup Q ∼= Cq. Thus P, Q C G and P ∩Q = 1, and Lagrange’s Theorem shows that PQ = G, and so G ∼= P × Q ∼= Cpq , by the product recognition theorem. This shows that all groups of order 15, 33, ... are cyclic.

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More is known: No group of order pqr is simple, where p, q and r are distinct primes (HW; counting argument as in 2i). Similarly groups of

2order p q and pα (for any α > 1) are never simple. Much deeper results: α βBurnside Theorem |G| = p q =⇒ G is not simple.

Feit-Thompson Theorem |G| odd (but not prime) =⇒ G is not simple

Proofs of Sylow Theorems (using group actions)

Let G × X → X, (g, x) 7→ g · x be an action of a group G on a set X. For any x ∈ X, define the orbit of x to be

Gx = {g · x | g ∈ G} ⊂ X

and the stabilizer (or isotropy subgroup) of x to be

Gx = {G ∈ G | g · x = x} < G.

(note the subscript). Call x a fixed point of the action if g · x = x for all g ∈ G, or equivalently Gx = {x}, or Gx = G. Let XG = {all fixed points}. The most important theorem in the subject is:

Orbit Stabilizer Theorem (OST) If G and X are finite, then the size of any orbit Gx is equal to the index of the corresponding stabilizer Gx. In symbols |Gx| = |G : Gx|, or equivalently |G| = |Gx||Gx|.

Proof The function G/Gx → Gx, gGx 7→ g ·x is a well defined bijection: −1gGx = hGx ⇐⇒ g−1h ∈ Gx ⇐⇒ g · h · x = x ⇐⇒ h · x = g · x.

−1Examples 1i Let G act on itself by conjugation, g · x = gxg . Then Gx = C(x) (the conjugacy class of x), Gx = Z(x) (the centralizer of x) and GG = Z(G) (the center of G). The OST says

|C(x)| = |G : Z(x)|.

2iLet G act on the set X of all its subgroups by conjugation, g · S = gSg−1 . Then GS = C(S) = {conjugates of S}, GS = N(S) (the normalizer of S) and XG = {normal subgroups of G}. The OST says

|C(S)| = |G : N(S)|.

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Now observe that the orbits of an action of G on X partition X (since Gx ∩ Gy 6= ∅ =⇒ g · x = h · y for some g, h ∈ G =⇒ k · y = (kh−1g) · x =⇒ Gy ⊂ Gx, and similarly Gx ⊂ Gy, so Gx = Gy). Thus X

|X| = |Gxi|

where the sum is over a set of representatives xi, one chosen from each orbit. Since |Gx| = |G : Gx| by the OST, the right hand side can be rewritten asP P |G : Gxi | = |XG| + |G : Gxi | where the last sum is only over those xi

that are not fixed points. This gives the class equation (for finite X and G) X |X| = |XG| + |G : Gxi |

summed over representatives xi of the non-trivial orbits. In the special case when G acts on itself by conjugation, as in Example 1i above, this reads X

|G| = |Z(G)| + |G : Z(xi)|

summed over representatives of the non-trivial conjugacy classes.

One very useful consequence of the general class equation, which is the key to our proof of Sylow’s Theorems, is

Lemma If a p-group P acts on a finite set X, then |X| ≡ |XP | (mod p). P Proof The class equation says |X| = |XP | + |P : Hi|, where the Hi

are proper subgps of P , and each |P : Hi| ≡ 0 (mod p) since P is a p-group, so |X| ≡ |XP | (mod p).

Corollary The center Z of any nontrivial p-group P is nontrivial.†

Proof Let P act on itself by conjugation. Then the fixed point set is Z, which by the lemma has order divisible by p, so cannot be trivial.

Now we are ready to prove Sylow’s Theorems.

†It follows that if |P | = p 2 , then P is abelian. Indeed |G/P | = 1 or p by the Corollary, and so P/Z is cyclic, generated say by some xZ. But then any element in P is of the form

k k ` k+` ` k x u for some k ∈ Z, u ∈ Z and so P is abelian: ∀u, v ∈ Z, x ux v = x uv = x vx u.

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Proof of Sylow I The result is obvious if G is the trivial group, so we αsuppose G nontrivial and induct on its order |G| = p m (with p - m)

assuming the result for groups of smaller order. Set Z = center of G. Case 1: p divides |Z|. Then it follows easily from the structure theorem

for finite abelian groups that Z has a subgroup P of order p =⇒ |G/Z| = α−1 α−1p m =⇒ (by induction) G/P has subgroups of order 1, p . . . , p =⇒

α(by the correspondence theorem) G has subgroups of order 1, p, . . . , p . Case 2: p does not divide |Z|. Then by the class equation, p does not

divide the index of the centralizer Z(x) of some x 6∈ Z =⇒ pα||Z(x)| =⇒ (by αinduction) Z(x) has subgroups of order 1, p, . . . , p =⇒ G does as well.

Proof of Sylow II For any two Sylow p-subgroups P and Q of G, let P act by left multiplication on G/Q. Since |G/Q| = m 6≡ 0 (mod p), it follows from the lemma that ∃ a fixed point xQ, i.e. for all g ∈ P , have

gxQ = xQ =⇒ gx ∈ xQ =⇒ g ∈ xQx−1 .

Thus P ⊂ xQx−1 , and so they are equal since they have the same order.

Note : This proof shows that any p-subgroup of G is a subset of some Sylow p-subgroup.

Proof of Sylow III Let X = {Sylow p-subgroups of G}, so

np = |X|.

Choose P ∈ X (by Sylow I) so X = C(P ), the set of all subgroups conjugate to P (by Sylow II). Thus |X| = |G : N(P )| (see example 2i on page 40), which is a divisor of m = |G : P | = |G : N(P )||N(P ) : P |, i.e. np|m. To see np ≡ 1 (mod p), consider the action of P on X by conjugation.

Since np|m, we have np 6≡ 0 (mod p) =⇒ ∃ a fixed point Q ∈ X, i.e. P ⊂ N(Q). But this forces P = Q, since Q C N(Q) (by definition) and so Q is the only Sylow p-subgroup of N(Q). Thus P is the unique fixed point of this action, so np ≡ 1 (mod p) by the lemma.

Exercises 1iFind all orders n < 100 for which Sylow III applies directly (without extra counting arguments as above) to produce a normal Sylow subgroup. (Answer: all except n = 12, 24, 30, 36, 48, 56, 60, 72, 80, 90, 96)

2i Show that S4 (of order 24) has no normal Sylow subgroups, but that every group of order less than 24 has at least one such subgroup.

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III Rings

§1. Basic Concepts

Definition A ring is a set R with two binary operations + and · satisfying ai (R, +) is an abelian group (with identity 0),

bi (R, ·) is a semigroup, and

ci · is distributive over + (on both sides)

Make sure you know what this entails (e.g. distributivity says that r(s+t) = rs + rt and (s + t)r = sr + tr, ∀r, s, t ∈ R). Say R is a ring with 1 if ∃1 ∈ R such that 1r = r1 = r (∀r ∈ R), and R is commutative if rs = sr (∀r, s ∈ R). A subset S ⊂ R is a subring of R, denoted S < R, if 0 ∈ S and

s, t ∈ S =⇒ −s, s + t and st ∈ S. Can show (as in group theory) that S < R ⇐⇒ S is nonempty and closed under subtraction and multiplication.

Examples 1i The trivial ring R = {0} is commutative with 1 = 0.

2iZ < Q < R < C < H are all rings with 1, and all but the last are comm.

3iZn is a commutative ring with 1, for any n ∈ N.

4iThe set Z[i] = {a + bi | a, b ∈ Z} of Gaussian integers is a subring of C. Similarly

Z[ζ] := {a + bζ | a, b ∈ Z} < C

2πi/3where ζ = e ∈ C; this ring will play a special role in the proof below of Fermat’s last theorem for n = 3. (Draw “pictures” of these rings as square/triangular lattice points in the plane)

5i(new rings from old) For any rings R and S have

(a) the product R × S with component-wise operations

(b) matrix rings Mn(R) of n×n matrices with entries in R with the usual addition and multiplication of matrices P n(c) polynomial rings R[x] = { i | ri ∈ R} with the usual addition i=0 rix and multiplication of polynomials, and in more variables R[x, y], R[x, y, z], etc.

(d) power series rings R[[x]] = { P∞ i | ri ∈ R}, R[[x, y]], etc.i=0 rix

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P n(e) group rings RG = { =0 rigi | ri ∈ R, gi ∈ G} of G (any group) over P P P n n nR, with the operations = (ri + si)gi andP P Pi=0 rigi + i=0 sigi i=0 n n n( i=0 rigi)( j=0 sj gj ) = i,j=0(risj )gigj

(f) function rings The set [X, R] of all functions from a set X to R under the operations (f + g)(x) = f(x) + g(x) and (fg)(x) = f(x)g(x).

Remark Many familiar properties of Z generalize to all rings, for example ai 0r = 0, bi (−r)s = r(−s) = −(rs), ci −1 · r = −r (in rings with 1); see the text for proofs. But not all, e.g. commutativity. Also, it is not true in general (even in commutative rings) that

rs = 0 =⇒ r or s = 0

e.g. 2 · 2 = 0 in Z4. Commutative rings with 1 6= 0 where this holds are called integral domains (more on this below).

Definition A function f : R → S, where R and S are rings, is called a ring homomorphism if

f(r + s) = f(r) + f(s) and f(rs) = f(r)f(s)

for all r, s ∈ R. If R and S are rings with identity, often also require that f(1) = 1. The kernel of f is defined by ker(f) = f−1(0) (not f−1(1)!) and the image im(f) is defined in the usual way. Both are subrings (of R and S respectively). Have the usual criterion f is 1-1 ⇐⇒ ker(f) = {0}.

Example 1i f : Z → Zn, f(k) = k is an epimorphism with kernel nZ.

2ig : Z → Zm × Zn, g(k) = (k, k) (e.g. if m = 5, n = 9 then f(13) = (3, 4)) is a homomorphism with kernel `Z, where ` is the lcm of m and n.

§2. Ideals and Quotient Rings

Definition A subring A of a ring R is an ideal in R, denoted A C R, if a ∈ A, r ∈ R =⇒ ra and ar ∈ A, i.e. A is closed under multiplication on the left or right by arbitrary elements of R.

ideals are the ring theory analogue of normal subgroups in group theory

Examples 0i Every ring R has the trivial ideals {0} and R; all other ideals are called proper.

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1inZ C Z for any n (these are the only ideals in Z).

2iLet R be a commutative ring with 1 and a ∈ R. Then

aR = {ar | r ∈ R} (also denoted hai)

is an ideal in R containing a (verify this) called the principal ideal generated by a. An ideal J C R is called a principal ideal if J = hai for some a ∈ R.

3iLet f : R → S be a ring morphism. Then ker(f) C R (verify this).

Definition Given an ideal J in a ring R, define the quotient ring R/J to be the set of cosets {r + J | r ∈ R} with operations + and · defined in the obvious way

(r + J) + (s + J) = (r + s) + J and (r + J)(s + J) = (rs) + J.

These operations are well-defined (e.g. for ·: if r + J = r0 + J , i.e. r − r0 ∈ J , 0 0then rs − r s = (r − r0)s ∈ J , so rs + J = r s + J ; similarly indep of the

choice of rep for s + J). They make R/J into a ring; the additive identity is 0 + J = J and the negative of r + J is (−r) + J .

There’s a universal property, as for gps, and isomorphism theorems, e.g.:

First Isomorphism Theorem If f : R → S is a ring homomorphism, then R/ ker(f) ∼= im(f).

Examples 1iUsing f : Z → Zn of example 1iin §1, have Z/nZ ∼= Zn.

2iUsing g : Z → Zm × Zn of example 2iin §1, have im(g) ∼= Z/`Z, where ` =lcm(m, n). In particular, if m and n are relatively prime, so ` = mn, then (counting elements) we see that g induces an isomorphism

Zmn =∼ Zm × Zn if (m, n) = 1

mapping k to (k, k). This =⇒ the “Chinese Remainder Theorem” that states that for any a, b ∈ Z, there exists x ∈ Z satisfying the system of congruences

x ≡ a (mod m) x ≡ b (mod n)

and all other solutions are of the form x+kmn for some k ∈ Z. For example, if m = 5, n = 9 and a = 2, b = 8, then have the soln set {17 + 45k | k ∈ Z}.

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§3. Integral Domains and Fields

Throughout this section, assume R is a commutative ring with 1 6= 0.

Definition A nonzero element a ∈ R is ai a zero divisor if ∃ b ∈ R − 0 such that ab = 0 bi a unit if ∃ b ∈ R such that ab = 1 (Note that b is unique, if it exists, and is denoted a−1.) Two elements a, b ∈ R are associates, written a ∼ b, if a = bu for some unit u. Set R◦ = {zero divisors in R} and R• = {units in R}. Then

R◦ ∩ R• = ∅

−10since a ∈ R◦ ∩ R• =⇒ 0 = ab for some b =6 0 =⇒ b = a−1ab = a = 0 which is a contradiction.

Definition‡ R is called an integral domain (or just a domain) if it has no zero divisors (i.e. R◦ = ∅), and a field if all nonzero elements are units (i.e. R• = R − {0}).

Clearly every field is a domain (since R◦ ∩ R• = ∅) but not conversely (e.g. Z is a domain but not a field). There is a useful characterization of fields in terms of ideals:

Lemma† R is a field ⇐⇒ R has no proper ideals.

Proof (=⇒) For any nonzero ideal J , choose a 6= 0 in J . Then for any −1r ∈ R have r = ra a ∈ J , so J = R. (⇐=) For any a 6= 0 in R, the

principal ideal aR 6= 0, so by hypothesis aR = R. Thus ∃r ∈ R such that ar = 1 so a is a unit. Thus R is a field.

There are two special kinds of ideals that relate to these notions:

Definition† Let J C R, J =6 R. Then J is prime if ab ∈ J =⇒ a ∈ J or b ∈ J , and J is maximal if J ⊂ K C R =⇒ K = J or K = R.

Exercise R is a domain ⇐⇒ {0} is a prime ideal.

Theorem Let R be a commutative ring with 1 6= 0 and J C R. Then ai J is prime ⇐⇒ R/J is a domain bi J is maximal ⇐⇒ R/J is a field.

Proof ai HW bi By the correspondence theorem J is maximal ⇐⇒ R/J has no proper ideals, and this is equivalent to R/J being a field by the previous lemma.

‡Recall that we are assuming that R is a commutative ring with 1 6= 0.

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Examples 1inZ C Z is prime ⇐⇒ maximal ⇐⇒ n is a prime number.

2i In general, maximal ideals are always prime (HW) but not conversely. For example hxi C Z[x] (consisting of all polynomials with 0 constant term) is prime but not maximal: Z[x]/hxi ∼= Z (and now apply the theorem).

Coda

Fermat’s Last Theorem (1637, proved by A. Wiles in 1995) For n ≥ 3, the equation

n n x n + y = z

has no non-zero integral solutions x, y, z. (Call this FLTn for any given n.)

Remarks 1i FLTn =⇒ FLTkn, since any solution x, y, z for kn would k kgive the solution x , y , zk for n. Fermat knew a proof for n = 4, and so this

reduces the proof to the case of odd primes, i.e. it suffices to prove that

p pxp + y = z

has no non-zero integral solutions x, y, z for any odd prime p.

2iA solution x, y, z is primitive if x, y and z are nonzero and have no common factor. Enough to prove 6 ∃ primitive solutions (since any soln cx, cy, cz gives another x, y, z). Note x, y, z primitive solution =⇒ pairwise relatively prime.

3iProof for p = 3 (which we give below) was known to Euler, for p = 5 to Dirichlet & Legendre, for p = 7 to Gabriel Lame. Indeed Lame and Cauchy (independently) thought they had it all. Their approach was to show 6 ∃ any solutions in the bigger ring

Λp := Z[ζp] = {polys in ζp with Z coeffs}

where ζp = e2πi/n; their mistake, uncovered by Kummer, was to assume that Λp has unique factorization into primes (as Z does) which in fact it does only for p ≤ 19. Kummer’s work led him to introduce “ideals” which then led to modern ring theory.

Proof (for n = 3) Set ζ = ζ3 and Λ = Λ3 = Z[ζ]. We will show that for any unit u ∈ Λ• ,

3 3 x 3 + y = uz

has no primitive solutions in Λ (FLT is the case u = 1). Note that

Λ• = {±1, ±ζ, ±ζ2}

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as seen by inspection (since these are the elements of complex norm 1 in Λ, and all other elements have norm > 1). It is a fact (which we won’t prove here) that any nonzero, nonunit in

Λ can be factored uniquely (up to order and replacement by associates) as a product of primes (nonunits whose only divisors are their associates and√ units). One such prime is p := ζ − 1. Why is p prime? Well, |p| = 3 which is the smallest norm achieved by a nonunit in Λ, so if p = ab then |p| = |a||b| =⇒ |a| or |b| = 1, i.e. a or b ∈ Λ• .

2Remarks 1i3 is not prime in Λ, but factors as 3 = p u, where u = −ζ2 . 3Thus p2|3 (also written 3 ≡ 2 0) but p - 3 (3 6≡ 3 0).p p

2iEach a ∈ Λ can be written uniquely in the form

a = a + rp

for some r ∈ Λ where a = 0, 1 or −1. This defines an isomorphism

Λ/pΛ → Z3 , a + pΛ 7→ a

To show this, write a as a poly w/ integral coeff in ζ, and thus in p by substituting p + 1 for ζ. Now reduce the constant term to 0 or ±1 using 1i

For example u = ±1 for any u ∈ Λ•; in particular ζ = ζ2 = 1.

3 33i For any a ∈ Λ, have a ≡ 3 a. Proof: a = a + rp, by 2i, so a = p 2 2 3a3 + 3a rp + 3ar p2 + r p3 ≡ 3 a3 (by 1i) = a.p

Now suppose xo, yo, zo is a primitive solution to

3 3 x 3 + y = uz

for some unit u. This is Fermat’s equation when u = 1, but as we shall see, it is easier to prove simultaneously that none of these six equations (for the various units u) have solutions.

Case 1 p - xoyozo. Then by the remarks above, xo, y , zo = ±1 =⇒o 3 3 30 = x + y − uz ≡ 3 xo + y ± zo = ±1 or ±3 6≡ 3 0 =⇒⇐=.o o o p o p

Case 2 p | xoyozo. Then p divides exactly one of xo, yo, zo, since they are pairwise relatively prime.

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Case 2a Suppose p | zo. Then xo = −y = ±1, so without loss of gener-o ality, xo = rp + 1 and yo = sp − 1 for suitable r, s ∈ Λ.

Let k be the largest natural number for which pk | zo, called the p-order of the solution. A simple calculation (reducing mod p4) shows that in fact k ≥ 2.† We assume that the solution was chosen so that k is minimal.

Now (and this is the magic!) define

xo + yo ζxo + ζ2yo ζ2xo + ζyo a = b = c = .

p p p

The numerators are all divisible by p (e.g. for b, ζxo + ζ2yo = xo + y = 0)o and so a, b, c ∈ Λ. A straightforward argument, using the fact that ζ3 = 1 and 1 + ζ + ζ2 = 0, shows

1ia + b + c = 0 3 32iabc = (x + y )/p3 = u(zo/p)3 o o

3ia, b, c are pairwise relatively prime, i.e. have no common prime factors (to see this, note that xo, yo are linearly related to any pair of a, b, c, and so a common factor for such a pair would yield one for xo, yo).

2iand 3ishow that each of a, b, c is a unit times a cube, and that these 3 3 3cubes are relatively prime. It follows from 1ithat ∃ x1, y1, z1 with x1, y1, z1 kassociates of a, b, c, in some order, such that p - x1, p - y1, p | z1, p - z1,

and 3 3 3 x1 + vy1 + wz = 01

for suitable units v, w. It is easy to check that v = ±1 (since the left hand 3side is congruent mod p to x1 + vy1 = ±1 ± v =⇒ v ≡ 3 ±1 =⇒ v = ±1)p

and so this gives a new solution x1, ±y1, z1 of smaller p-order to one of the original equations, contradicting the minimality of k.

Case 2b Suppose p | xo (or yo). Then uz ≡p3 x + y =⇒ u ≡p3 ±1 =⇒ 3u = ±1 =⇒ (−yo)3 + (±zo)3 = x , which is handled in case 2a.o

Thus Fermat’s Last Theorem for n = 3 is proved.

† 3 3 3 3 3 x0 + y0 = (rp + 1)3 + (sp − 1)3 ≡p4 r + s 3 − ζ2(r + s))p ≡p4 (r + s − (r + s))p = 0 =⇒ z ≡ 2 0.p

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